Question: There is an angle $\theta$ in the range $0^\circ < \theta < 45^\circ$ which satisfies
\[\tan \theta + \tan 2 \theta + \tan 3 \theta = 0.\]Calculate $\tan \theta$ for this angle.
Solution: Let $t = \tan \theta.$  Then $\tan 2 \theta = \frac{2t}{1 - t^2}$ and $\tan 3 \theta = \frac{3t - t^3}{1 - 3t^2},$ so
\[t + \frac{2t}{1 - t^2} + \frac{3t - t^3}{1 - 3t^2} = 0.\]This simplifies to $4t^5 - 14t^3 + 6t = 0.$  This factors as $2t(2t^2 - 1)(t^2 - 3) = 0.$

Since $0^\circ < \theta < 45^\circ,$ $0 < t < 1.$  The only solution in this interval is $t = \boxed{\frac{1}{\sqrt{2}}}.$